6.3 - Substitution
As I mentioned before, there is no such thing as a product rule or a quotient rule for integrals.
This means we need another way to solve that stu▯, which in short, means extra work. Sorry.
Tough luck. So, here’s the idea behind substitution:
Think back to derivatives and the chain rule.
d 0 0
dx F(g(x)) = F (g(x))g (x)
Well, remember, the integral is just everything backwards, so
F (g(x))g (x) dx = F(g(x)) + C
But recall from before that F (x) = f(x), so this all comes down to
f(g(x))g (x) dx = F(g(x)) + C
Well, that’s a bit confusing! So, let’s make u = g(x). Then= g (x), or equivalently,
du = g (x)dx (think back to di▯erentials and how we treated dx like a variable there!). This means
f(u)du = F(u) + C
This is what’s known as u-substitution.
2x 1 + x dx
So how do we tackle this problem? The idea is that we need to let u be something. You can
certainly do trial and error, but the catch is you have to cancel out all the x’s. Once you do a
u-substitution, there can only be u’s and constants left. This means when we choose u, we need to
have du cancel out the other x’s. Here, we are going to let u = 1 + x . This means du = 2x dx.
Solve for dx and you get dx =du. Now substitute.
Z p Z Z
2 p du p
2x 1 + x dx = 2x u 2x = u du
Ah, much easier to integrate! Z
p u du = 2u3=2+ C
DO NOT leave the answer in terms of u! You have to put it back in terms of x, because that’s
what we started with!
= (1 + x )3=2+ C
1 So how do we go about choosing the right u? There’s no sure-▯re way to always know what to do,
but here a suggestion that might help:
Look for composition of functions f(g(x)) and try letting u = g(x). Examples: 2x(x + 5)
results in u = x + 5. Or, perhaps sin(x)sin(cos(x)) will have u = cos(x). In other words, let u be
the stu▯ under a square root or the polynomial inside) for some n. Again, this isn’t always
going to work, but 9 times out of 10 it probably will. Remember, you have to be able to get rid of
everything that has an x in it!
x cos(x + 2) dx
let u = x + 2. Then du = 4x dx. So our integral turns into
Z Z 4
3 du 1 sin(u) sin(x + 2)
x cos(u) 4x3 = 4cos(u) du = 4 + C = 4 + C
let u = 5x. Then du = 5 dx and our integral turns into
1 ▯cos(u) ▯cos(5x)
sin(u) du = + C = + C
5 5 5
Notice what happens here. When there’s no other function you have to cancel out the derivative
with (ie your u = cx for some constant c), then when you take the integral your answer ends up as
the integral of the outside function divided by the constant. If this is confusing, forget that i just
mentioned this and just continue using u-substitution. It takes a little longer, that’s all.
Now, you don’t always have to choose u to be the inside function. A lot of times, you can let u be
something else and it still works out in the end.
2x + 1 dx
let’s solve this two di▯erent ways.
Solution: (1) let u = 2x + 1. Then du = 2 dx and we end up with
Z p 3=2
u du = 2 u + C = 1 (2x + 1)=2+ C
2 3 2 3
p p 2
Solution: (2) let u = 2x + 1. Then du = 2 2x+1dx. So, we get
Z p p Z 1 1
2x + 1 2x + 1 du = u du = u + C = (2x + 1)=2+ C
So, you don’t have to always have u be the inside function in a composition of functions. In fact,
letting u be the inside function doesn’t always work!
2 Example: Z
1 + x4dx
if we were to try